Question

A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 2% margin of error at a 97.5% confidence level, what size of sample is needed?Give your answer in whole people.

Answer 1

Answer:

The sample size is  $n = 3139$    

Step-by-step explanation:

From the question we are told that

   The margin of error is  $E = 0.02$

From the question we are told the confidence level is  97.5% , hence the level of significance is    

      $\alpha = (100 - 97.5% ) \%$

=>   $\alpha = 0.025$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 2.241$

Generally we will assume the sample proportion to be $\^ p = 0.5$

Generally the sample size is mathematically represented as  

    $n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=>  $n = [\frac{2.241}{0.02} ]^2 *0.5 (1 - 0.5 )$

=>  $n = 3139$